\(a,\text{ Để A }\in\text{ Z }\Leftrightarrow\text{ }\left(n+1\right)\inƯ\left(2\right)\)
\(\text{Mà }Ư\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\text{Do đó:}\) \(n+1=1\Leftrightarrow n=0\)
\(\text{hoặc }n+1=-1\Leftrightarrow n=-2\)
\(\text{hoặc }n+1=2\Leftrightarrow n=1\)
\(\text{hoặc }n+1=-2\Leftrightarrow n=-3\)
\(\text{Vậy: A }\in Z\Leftrightarrow n=\left\{0;-2;1;-3\right\}.\)
\(\text{a) Để B}\in Z\Leftrightarrow n-2\inƯ\left(3\right)\)
\(\text{Mà }Ư\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\text{Do đó: }n-2=1\Leftrightarrow n=3\)
\(\text{hoặc }n-2=-1\Leftrightarrow n=1\)
\(\text{hoặc }n-2=3\Leftrightarrow n=5\)
\(\text{hoặc }n-2=-3\Leftrightarrow n=-1\)
\(\text{Vậy: B}\in Z\Leftrightarrow n=\left\{3;1;5;-1\right\}.\)
ĐK n≠-1
a, ta có A=\(\frac{2}{n+1}\) để A∈Z ta có
2⋮(n+1)
=> n+1∈Ư(2)\(\left\{1;-1;2;-2\right\}\)
n+1=1 =>n=0 tm
n+1=-1 =>n=-2 tm
n+1=2 =>n=1 tm
n+1=-2 =>n=-3 tm
Vậy vs n=0;-2;1;-3 thì A∈Z
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